Does the evil cast get trumped by the evil compiler?

Question

This is not academic code or a hypothetical quesiton. The original problem was converting code from HP11 to HP1123 Itanium. Basically it boils down to a compile error on HP1123 Itanium. It has me really scratching my head when reproducing it on Windows for study. I have stripped all but the most basic aspects... You may have to press control D to exit a console window if you run it as is:

#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    char blah[6];
    const int IAMCONST = 3;
    int *pTOCONST;
    pTOCONST = (int *)  &IAMCONST;
    (*pTOCONST) = 7;
    printf("IAMCONST %d \n",IAMCONST);
    printf("WHATISPOINTEDAT %d \n",(*pTOCONST));
    printf("Address of IAMCONST %x   pTOCONST %x\n",&IAMCONST, (pTOCONST));
    cin >> blah;    
    return 0;
}

Here is the output

IAMCONST 3
WHATISPOINTEDAT 7
Address of IAMCONST 35f9f0   pTOCONST 35f9f0

All I can say is what the heck? Is it undefined to do this? It is the most counterintuitive thing I have seen for such a simple example.

Update:

Indeed after searching for a while the Menu Debug >> Windows >> Disassembly had exactly the optimization that was described below.

    printf("IAMCONST %d \n",IAMCONST);
0024360E  mov         esi,esp 
00243610  push        3    
00243612  push        offset string "IAMCONST %d \n" (2458D0h) 
00243617  call        dword ptr [__imp__printf (248338h)] 
0024361D  add         esp,8 
00243620  cmp         esi,esp 
00243622  call        @ILT+325(__RTC_CheckEsp) (24114Ah)

Thank you all!

Answer 1

The constant value IAMCONST is being inlined into the printf call.

What you're doing is at best wrong and in all likelihood is undefined by the C++ standard. My guess is that the C++ standard leaves the compiler free to inline a const primitive which is local to a function declaration. The reason being that the value should not be able to change.

Then again, it's C++ where should and can are very different words.

Answer 2

Looks like the compiler is optimizing

printf("IAMCONST %d \n",IAMCONST);

into

printf("IAMCONST %d \n",3);

since you said that IAMCONST is a const int.

But since you're taking the address of IAMCONST, it has to actually be located on the stack somewhere, and the constness can't be enforced, so the memory at that location (*pTOCONST) is mutable after all.

In short: you casted away the constness, don't do that. Poor, defenseless C...

Addendum

Using GCC for x86, with -O0 (no optimizations), the generated assembly

main:
    leal 4(%esp), %ecx
    andl $-16, %esp
    pushl -4(%ecx)
    pushl %ebp
    movl %esp, %ebp
    pushl %ecx
    subl $36, %esp
    movl $3, -12(%ebp)
    leal -12(%ebp), %eax
    movl %eax, -8(%ebp)
    movl -8(%ebp), %eax
    movl $7, (%eax)
    movl -12(%ebp), %eax
    movl %eax, 4(%esp)
    movl $.LC0, (%esp)
    call printf
    movl -8(%ebp), %eax
    movl (%eax), %eax
    movl %eax, 4(%esp)
    movl $.LC1, (%esp)
    call printf

copies from *(bp-12) on the stack to printf's arguments. However, using -O1 (as well as -Os, -O2, -O3, and other optimization levels),

main:
    leal 4(%esp), %ecx
    andl $-16, %esp
    pushl -4(%ecx)
    pushl %ebp
    movl %esp, %ebp
    pushl %ecx
    subl $20, %esp
    movl $3, 4(%esp)
    movl $.LC0, (%esp)
    call printf
    movl $7, 4(%esp)
    movl $.LC1, (%esp)
    call printf

you can clearly see that the constant 3 is used instead.

If you are using Visual Studio's CL.EXE, /Od disables optimization. This varies from compiler to compiler.

Be warned that the C specification allows the C compiler to assume that the target of any int * pointer never overlaps the memory location of a const int, so you really shouldn't be doing this at all if you want predictable behavior.

Answer 3

You are lucky the compiler is doing the optimization. An alternative treatment would be to place the const integer into read-only memory, whereupon trying to modify the value would cause a core dump.

Answer 4

Writing to a const object through a cast that removes the const is undefined behavior - so at the point where you do this:

(*pTOCONST) = 7;

all bets are off.

From the C++ standard 7.1.5.1 (The cv-qualifiers):

Except that any class member declared mutable (7.1.1) can be modified, any attempt to modify a const object during its lifetime (3.8) results in undefined behavior.

Because of this, the compiler is free to assume that the value of IAMCONST will not change, so it can optimize away the access to the actual storage. In fact, if the address of the const object is never taken, the compiler may eliminate the storage for the object altogether.

Also note that (again in 7.1.5.1):

A variable of non-volatile const-qualified integral or enumeration type initialized by an integral constant expression can be used in integral constant expressions (5.19).

Which means IAMCONST can be used in compile-time constant expressions (ie., to provide a value for an enumeration or the size of an array). What would it even mean to change that at runtime?

Answer 5

It doesn't matter if the compiler optimizes or not. You asked for trouble and you're lucky you got the trouble yourself instead of waiting for customers to report it to you.

"All I can say is what the heck? Is it undefined to do this? It is the most counterintuitive thing I have seen for such a simple example."

If you really believe that then you need to switch to a language you can understand, or change professions. For the sake of yourself and your customers, stop using C or C++ or C#.

const int IAMCONST = 3;

You said it.

int *pTOCONST;
pTOCONST = (int *)  &IAMCONST;

Guess why the compiler complained if you omitted your evil cast. The compiler might have been telling the truth before you lied to it.

"Does the evil cast get trumped by the evil compiler?"

No. The evil cast gets trumped by itself. Whether or not your compiler tried to tell you the truth, the compiler was not evil.